Integration by substitution Questions in English

(A) Let $t = 1 + \sin 6x$.
Then,differentiating with respect to $x$,we get $dt = 6 \cos 6x \, dx$,which implies $\cos 6x \, dx = \frac{1}{6} \, dt$.
Substituting these into the integral:
$\int \cos 6x \sqrt{1+\sin 6x} \, dx = \int \sqrt{t} \cdot \frac{1}{6} \, dt$
$= \frac{1}{6} \int t^{\frac{1}{2}} \, dt$
$= \frac{1}{6} \left( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right) + C$
$= \frac{1}{6} \cdot \frac{2}{3} t^{\frac{3}{2}} + C$
$= \frac{1}{9} (1 + \sin 6x)^{\frac{3}{2}} + C$

(A) We have $\int \frac{(x^{4}-x)^{\frac{1}{4}}}{x^{5}} dx = \int \frac{[x^{4}(1-\frac{1}{x^{3}})]^{\frac{1}{4}}}{x^{5}} dx = \int \frac{x(1-\frac{1}{x^{3}})^{\frac{1}{4}}}{x^{5}} dx = \int \frac{(1-\frac{1}{x^{3}})^{\frac{1}{4}}}{x^{4}} dx$.
Let $t = 1 - \frac{1}{x^{3}} = 1 - x^{-3}$.
Then $dt = -(-3)x^{-4} dx = \frac{3}{x^{4}} dx$,which implies $\frac{dx}{x^{4}} = \frac{dt}{3}$.
Substituting these into the integral,we get $\int \frac{t^{\frac{1}{4}}}{3} dt = \frac{1}{3} \int t^{\frac{1}{4}} dt$.
Integrating,we get $\frac{1}{3} \times \frac{t^{\frac{5}{4}}}{\frac{5}{4}} + C = \frac{1}{3} \times \frac{4}{5} t^{\frac{5}{4}} + C = \frac{4}{15} t^{\frac{5}{4}} + C$.
Substituting back $t = 1 - \frac{1}{x^{3}}$,we get $\frac{4}{15}(1-\frac{1}{x^{3}})^{\frac{5}{4}} + C$.

(NONE) Let $I = \int \frac{\sin 2x \cos 2x \, dx}{\sqrt{9-\cos^{4}(2x)}}$.
Substitute $t = \cos^{2}(2x)$.
Then,$dt = 2 \cos(2x) \cdot (-\sin(2x)) \cdot 2 \, dx = -4 \sin(2x) \cos(2x) \, dx$.
This implies $\sin(2x) \cos(2x) \, dx = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int \frac{-\frac{1}{4} dt}{\sqrt{9-t^{2}}} = -\frac{1}{4} \int \frac{dt}{\sqrt{3^{2}-t^{2}}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin^{-1}(\frac{x}{a}) + C$:
$I = -\frac{1}{4} \sin^{-1}\left(\frac{t}{3}\right) + C$.
Substituting $t = \cos^{2}(2x)$ back:
$I = -\frac{1}{4} \sin^{-1}\left(\frac{\cos^{2}(2x)}{3}\right) + C$.

Let $I = \int \frac{1}{x \sqrt{ax - x^{2}}} dx$.
Substitute $x = \frac{a}{t}$,then $dx = -\frac{a}{t^{2}} dt$.
Substituting these into the integral:
$I = \int \frac{1}{\frac{a}{t} \sqrt{a \cdot \frac{a}{t} - \left(\frac{a}{t}\right)^{2}}} \left(-\frac{a}{t^{2}} dt\right)$
$I = \int \frac{1}{\frac{a}{t} \sqrt{\frac{a^{2}}{t} - \frac{a^{2}}{t^{2}}}} \left(-\frac{a}{t^{2}} dt\right)$
$I = \int \frac{1}{\frac{a}{t} \cdot \frac{a}{t} \sqrt{t - 1}} \left(-\frac{a}{t^{2}} dt\right)$
$I = -\int \frac{1}{\sqrt{t - 1}} dt$
$I = -2 \sqrt{t - 1} + C$
Substituting $t = \frac{a}{x}$ back:
$I = -2 \sqrt{\frac{a}{x} - 1} + C = -2 \sqrt{\frac{a - x}{x}} + C$.

To integrate $I = \int \frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}} dx$,we manipulate the integrand.
Multiply and divide by $x^{-3}$:
$I = \int \frac{x^{-3}}{x^{2} x^{-3}\left(x^{4}+1\right)^{\frac{3}{4}}} dx = \int \frac{x^{-3}}{\left(x^{4}+1\right)^{\frac{3}{4}} x^{-1}} dx$
Rewrite the expression inside the integral:
$I = \int \frac{x^{-3}}{\left(x^{4}\left(1+\frac{1}{x^{4}}\right)\right)^{\frac{3}{4}}} dx = \int \frac{x^{-3}}{x^{4 \cdot \frac{3}{4}} \left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}} dx$
$I = \int \frac{x^{-3}}{x^{3} \left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}} dx = \int \frac{1}{x^{6}} \left(1+\frac{1}{x^{4}}\right)^{-\frac{3}{4}} dx$
Let $t = 1 + \frac{1}{x^{4}}$. Then $dt = -\frac{4}{x^{5}} dx$,which implies $\frac{1}{x^{5}} dx = -\frac{dt}{4}$.
Wait,let's re-evaluate the substitution:
$I = \int \frac{1}{x^{2} \cdot x^{3} (1 + x^{-4})^{3/4}} dx = \int x^{-5} (1 + x^{-4})^{-3/4} dx$
Let $u = 1 + x^{-4}$. Then $du = -4x^{-5} dx$,so $x^{-5} dx = -\frac{du}{4}$.
$I = -\frac{1}{4} \int u^{-3/4} du = -\frac{1}{4} \left( \frac{u^{1/4}}{1/4} \right) + C = -u^{1/4} + C$
Substituting back $u = 1 + \frac{1}{x^{4}}$:
$I = -\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}} + C$

Let $x = t^{6}$,then $dx = 6t^{5} dt$.
Substituting these into the integral:
$\int \frac{1}{x^{1/2} + x^{1/3}} dx = \int \frac{6t^{5}}{t^{3} + t^{2}} dt$
$= \int \frac{6t^{5}}{t^{2}(t + 1)} dt = 6 \int \frac{t^{3}}{t + 1} dt$
By performing polynomial division,$\frac{t^{3}}{t + 1} = t^{2} - t + 1 - \frac{1}{t + 1}$.
So,the integral becomes:
$6 \int (t^{2} - t + 1 - \frac{1}{t + 1}) dt = 6 [\frac{t^{3}}{3} - \frac{t^{2}}{2} + t - \log|t + 1|] + C$
$= 2t^{3} - 3t^{2} + 6t - 6 \log|t + 1| + C$
Since $t = x^{1/6}$,substituting back gives:
$= 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6 \log|x^{1/6} + 1| + C$

To integrate $I = \int \frac{\sin x}{\sin (x-a)} dx$,we use the substitution method.
Let $t = x-a$,which implies $x = t+a$ and $dx = dt$.
Substituting these into the integral:
$I = \int \frac{\sin (t+a)}{\sin t} dt$
Using the trigonometric identity $\sin (A+B) = \sin A \cos B + \cos A \sin B$:
$I = \int \frac{\sin t \cos a + \cos t \sin a}{\sin t} dt$
Dividing each term by $\sin t$:
$I = \int (\cos a + \cot t \sin a) dt$
Integrating with respect to $t$:
$I = \cos a \int dt + \sin a \int \cot t dt$
$I = t \cos a + \sin a \ln |\sin t| + C_1$
Substituting $t = x-a$ back into the expression:
$I = (x-a) \cos a + \sin a \ln |\sin (x-a)| + C_1$
$I = x \cos a - a \cos a + \sin a \ln |\sin (x-a)| + C_1$
Since $-a \cos a$ is a constant,we can combine it with $C_1$ to form a new constant $C$:
$I = x \cos a + \sin a \ln |\sin (x-a)| + C$

(B) To integrate the function $\int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} dx$,we use the method of substitution.
Let $\sin x = t$.
Then,differentiating both sides with respect to $x$,we get $\cos x dx = dt$.
Substituting these into the integral,we have:
$\int \frac{dt}{\sqrt{2^2 - t^2}}$.
Using the standard integration formula $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}(\frac{x}{a}) + C$,where $a = 2$ and $x = t$,we get:
$\sin^{-1}(\frac{t}{2}) + C$.
Substituting back $t = \sin x$,the final result is:
$\sin^{-1}(\frac{\sin x}{2}) + C$.

(A) To integrate $\int \frac{x^{3}}{\sqrt{1-x^{8}}} dx$,we use the method of substitution.
Let $t = x^{4}$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = 4x^{3}$,which implies $x^{3} dx = \frac{1}{4} dt$.
Substituting these into the integral:
$\int \frac{x^{3}}{\sqrt{1-(x^{4})^{2}}} dx = \int \frac{1}{\sqrt{1-t^{2}}} \cdot \frac{1}{4} dt$
$= \frac{1}{4} \int \frac{1}{\sqrt{1-t^{2}}} dt$
Using the standard integral formula $\int \frac{1}{\sqrt{1-t^{2}}} dt = \sin^{-1}(t) + C$,we get:
$= \frac{1}{4} \sin^{-1}(t) + C$
Substituting $t = x^{4}$ back into the expression:
$= \frac{1}{4} \sin^{-1}(x^{4}) + C$

(A) Given integral is $I = \int \cos ^{3} x e^{\log \sin x} dx$.
Using the property $e^{\log f(x)} = f(x)$,we have $e^{\log \sin x} = \sin x$.
So,the integral becomes $I = \int \cos ^{3} x \sin x dx$.
Let $\cos x = t$. Then,differentiating both sides with respect to $x$,we get $-\sin x dx = dt$,or $\sin x dx = -dt$.
Substituting these into the integral,we get $I = \int t^{3} (-dt) = -\int t^{3} dt$.
Integrating $t^{3}$ with respect to $t$,we get $I = -\frac{t^{4}}{4} + C$.
Substituting $t = \cos x$ back,we get $I = -\frac{\cos ^{4} x}{4} + C$.

Given integral is $I = \int e^{3 \log x}(x^{4}+1)^{-1} dx$.
Using the property of logarithms,$n \log x = \log x^{n}$,we have $e^{3 \log x} = e^{\log x^{3}} = x^{3}$.
Thus,the integral becomes $I = \int \frac{x^{3}}{x^{4}+1} dx$.
Let $t = x^{4}+1$. Then $dt = 4x^{3} dx$,which implies $x^{3} dx = \frac{dt}{4}$.
Substituting these into the integral,we get $I = \int \frac{1}{t} \cdot \frac{dt}{4} = \frac{1}{4} \int \frac{1}{t} dt$.
Integrating,we get $I = \frac{1}{4} \log |t| + C$.
Substituting back $t = x^{4}+1$,we get $I = \frac{1}{4} \log |x^{4}+1| + C$.
Since $x^{4}+1 > 0$ for all real $x$,we can write $I = \frac{1}{4} \log (x^{4}+1) + C$.

To integrate $I = \int f^{\prime}(ax+b)[f(ax+b)]^n \, dx$,we use the method of substitution.
Let $u = f(ax+b)$.
Then,differentiating with respect to $x$,we get $du = f^{\prime}(ax+b) \cdot a \, dx$,which implies $f^{\prime}(ax+b) \, dx = \frac{1}{a} \, du$.
Substituting these into the integral:
$I = \int u^n \cdot \frac{1}{a} \, du$
$I = \frac{1}{a} \int u^n \, du$
Using the power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$):
$I = \frac{1}{a} \cdot \frac{u^{n+1}}{n+1} + C$
Substituting $u = f(ax+b)$ back into the expression:
$I = \frac{[f(ax+b)]^{n+1}}{a(n+1)} + C$

Let $I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx$.
Put $x = \cos^2 \theta$,so $dx = -2 \sin \theta \cos \theta d\theta$.
$I = \int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} (-2 \sin \theta \cos \theta) d\theta$
$= -2 \int \sqrt{\frac{2 \sin^2 (\theta/2)}{2 \cos^2 (\theta/2)}} \sin \theta \cos \theta d\theta$
$= -2 \int \tan(\theta/2) \cdot (2 \sin(\theta/2) \cos(\theta/2)) \cos \theta d\theta$
$= -4 \int \sin^2(\theta/2) \cos \theta d\theta$
$= -4 \int \left(\frac{1-\cos \theta}{2}\right) \cos \theta d\theta$
$= -2 \int (\cos \theta - \cos^2 \theta) d\theta$
$= -2 \int \cos \theta d\theta + 2 \int \left(\frac{1+\cos 2\theta}{2}\right) d\theta$
$= -2 \sin \theta + \theta + \frac{\sin 2\theta}{2} + C$
$= -2 \sin \theta + \theta + \sin \theta \cos \theta + C$
Since $x = \cos^2 \theta$,$\cos \theta = \sqrt{x}$ and $\sin \theta = \sqrt{1-x}$.
$I = -2 \sqrt{1-x} + \cos^{-1}(\sqrt{x}) + \sqrt{1-x} \cdot \sqrt{x} + C$
$= \cos^{-1}(\sqrt{x}) - \sqrt{1-x} (2 - \sqrt{x}) + C$

Let $I = \int \frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}} dx$
$\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-\log x^{2}\right]}{x^{4}} = \frac{\sqrt{x^{2}+1}}{x^{4}} \log \left(\frac{x^{2}+1}{x^{2}}\right) = \frac{\sqrt{x^{2}+1}}{x^{4}} \log \left(1+\frac{1}{x^{2}}\right)$
$= \frac{1}{x^{3}} \sqrt{\frac{x^{2}+1}{x^{2}}} \log \left(1+\frac{1}{x^{2}}\right) = \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} \log \left(1+\frac{1}{x^{2}}\right)$
Let $1+\frac{1}{x^{2}} = t$. Then $-\frac{2}{x^{3}} dx = dt$,so $\frac{1}{x^{3}} dx = -\frac{1}{2} dt$.
$I = -\frac{1}{2} \int \sqrt{t} \log t \, dt = -\frac{1}{2} \int t^{\frac{1}{2}} \log t \, dt$
Using integration by parts: $\int u v \, dt = u \int v \, dt - \int (u' \int v \, dt) dt$
$I = -\frac{1}{2} \left[ \log t \cdot \frac{t^{3/2}}{3/2} - \int \frac{1}{t} \cdot \frac{t^{3/2}}{3/2} \, dt \right] = -\frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t - \frac{2}{3} \int t^{1/2} \, dt \right]$
$I = -\frac{1}{3} t^{3/2} \log t + \frac{1}{3} \cdot \frac{2}{3} t^{3/2} + C = -\frac{1}{3} t^{3/2} \log t + \frac{2}{9} t^{3/2} + C$
Substituting $t = 1+\frac{1}{x^{2}}$:
$I = -\frac{1}{3} \left(1+\frac{1}{x^{2}}\right)^{3/2} \left[ \log \left(1+\frac{1}{x^{2}}\right) - \frac{2}{3} \right] + C$

(A) Let $I = \int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{4 x^{2}-4 x+6}} d x$.
Note that $4x^2 - 4x + 6 = (2x-1)^2 + 5$.
So,$I = \int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{(2 x-1)^{2}+5}} d x$.
Let $u = \sqrt{(2 x-1)^{2}+5}$.
Then $u^2 = (2x-1)^2 + 5$.
Differentiating both sides with respect to $x$,we get $2u \frac{du}{dx} = 2(2x-1) \cdot 2 = 4(2x-1)$.
Thus,$(2x-1) dx = \frac{u}{2} du$.
Substituting these into the integral:
$I = \int \frac{\cos u}{u} \cdot \frac{u}{2} du = \frac{1}{2} \int \cos u du$.
$I = \frac{1}{2} \sin u + c$.
Substituting back $u = \sqrt{(2 x-1)^{2}+5}$,we get $I = \frac{1}{2} \sin \sqrt{(2 x-1)^{2}+5} + c$.

(C) Given $f(x) = \int \frac{5x^{8} + 7x^{6}}{(x^{2} + 1 + 2x^{7})^{2}} dx$.
Divide the numerator and denominator by $x^{14}$ inside the integral:
$f(x) = \int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + x^{-7} + 2)^{2}} dx$.
Let $t = x^{-5} + x^{-7} + 2$.
Then $dt = (-5x^{-6} - 7x^{-8}) dx$,which implies $-(5x^{-6} + 7x^{-8}) dx = dt$.
Substituting this into the integral:
$f(x) = \int -\frac{dt}{t^{2}} = \frac{1}{t} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C = \frac{x^{7}}{1 + x^{2} + 2x^{7}} + C$.
Given $f(0) = 0$,we find $C = 0$.
Thus,$f(x) = \frac{x^{7}}{x^{2} + 1 + 2x^{7}}$.
Evaluating at $x = 1$:
$f(1) = \frac{1^{7}}{1^{2} + 1 + 2(1)^{7}} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$.
Since $f(1) = \frac{1}{K}$,we have $K = 4$.

(B) Given integral: $I = \int \frac{e^{3 \log_{e} 2x} + 5e^{2 \log_{e} 2x}}{e^{4 \log_{e} x} + 5e^{3 \log_{e} x} - 7e^{2 \log_{e} x}} dx$
Using the property $e^{\log_{e} f(x)} = f(x)$,we have:
$I = \int \frac{(2x)^{3} + 5(2x)^{2}}{x^{4} + 5x^{3} - 7x^{2}} dx$
$I = \int \frac{8x^{3} + 20x^{2}}{x^{4} + 5x^{3} - 7x^{2}} dx$
Factor out $4x^{2}$ from the numerator and $x^{2}$ from the denominator:
$I = \int \frac{4x^{2}(2x + 5)}{x^{2}(x^{2} + 5x - 7)} dx$
$I = 4 \int \frac{2x + 5}{x^{2} + 5x - 7} dx$
Let $u = x^{2} + 5x - 7$,then $du = (2x + 5) dx$.
$I = 4 \int \frac{1}{u} du = 4 \log_{e} |u| + c$
$I = 4 \log_{e} |x^{2} + 5x - 7| + c$

(C) Let $I = \int \frac{dx}{(x-1)^{3/4}(x+2)^{5/4}}$.
Rewrite the integrand as: $I = \int \frac{dx}{\left(\frac{x+2}{x-1}\right)^{5/4} \cdot (x-1)^{3/4} \cdot (x-1)^{5/4}} = \int \frac{dx}{\left(\frac{x+2}{x-1}\right)^{5/4} \cdot (x-1)^2}$.
Let $t = \frac{x+2}{x-1}$. Then $dt = \frac{(x-1)(1) - (x+2)(1)}{(x-1)^2} dx = \frac{-3}{(x-1)^2} dx$,so $\frac{dx}{(x-1)^2} = -\frac{1}{3} dt$.
Substituting these into the integral: $I = \int \frac{-1/3}{t^{5/4}} dt = -\frac{1}{3} \int t^{-5/4} dt$.
Integrating: $I = -\frac{1}{3} \left( \frac{t^{-1/4}}{-1/4} \right) + C = \frac{4}{3} t^{-1/4} + C$.
Substituting back $t = \frac{x+2}{x-1}$: $I = \frac{4}{3} \left( \frac{x+2}{x-1} \right)^{-1/4} + C = \frac{4}{3} \left( \frac{x-1}{x+2} \right)^{1/4} + C$.

(A) Let $I = \int \frac{(x^8-x^2) dx}{(x^{12}+3x^6+1) \tan^{-1}(x^3+\frac{1}{x^3})}$.
Divide the numerator and denominator by $x^6$:
$I = \int \frac{(x^2 - x^{-4}) dx}{(x^6 + 3 + x^{-6}) \tan^{-1}(x^3 + x^{-3})}$.
Let $t = \tan^{-1}(x^3 + x^{-3})$.
Then $dt = \frac{1}{1 + (x^3 + x^{-3})^2} \cdot (3x^2 - 3x^{-4}) dx$.
$dt = \frac{3(x^2 - x^{-4})}{1 + x^6 + 2 + x^{-6}} dx = \frac{3(x^2 - x^{-4})}{x^6 + 3 + x^{-6}} dx$.
Thus,$\frac{(x^2 - x^{-4}) dx}{x^6 + 3 + x^{-6}} = \frac{1}{3} dt$.
Substituting this into the integral:
$I = \int \frac{1}{3t} dt = \frac{1}{3} \ln |t| + C$.
$I = \frac{1}{3} \ln |\tan^{-1}(x^3 + x^{-3})| + C = \ln |\tan^{-1}(x^3 + x^{-3})|^{1/3} + C$.
Therefore,the correct option is $A$.

(D) Given the integral $y(x) = \int \frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x \cos x} + \frac{\sin^3 x}{\cos x}} \, dx$.
Simplifying the integrand:
$y(x) = \int \frac{\frac{1+\sin^2 x}{\sin x}}{\frac{1+\sin^4 x}{\sin x \cos x}} \, dx = \int \frac{(1+\sin^2 x) \cos x}{1+\sin^4 x} \, dx$.
Let $\sin x = t$,then $\cos x \, dx = dt$.
$y(t) = \int \frac{1+t^2}{1+t^4} \, dt = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \, dt = \int \frac{1 + \frac{1}{t^2}}{(t - \frac{1}{t})^2 + 2} \, dt$.
Let $u = t - \frac{1}{t}$,then $du = (1 + \frac{1}{t^2}) \, dt$.
$y(u) = \int \frac{du}{u^2 + 2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t - \frac{1}{t}}{\sqrt{2}}\right) + C$.
As $x \rightarrow \frac{\pi}{2}$,$t \rightarrow 1$,so $u \rightarrow 0$.
Given $\lim_{x \rightarrow (\frac{\pi}{2})^-} y(x) = 0$,we have $\frac{1}{\sqrt{2}} \tan^{-1}(0) + C = 0 \implies C = 0$.
At $x = \frac{\pi}{4}$,$t = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
$u = \frac{1}{\sqrt{2}} - \sqrt{2} = \frac{1-2}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$.
Thus,$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{-1/\sqrt{2}}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} \tan^{-1}\left(-\frac{1}{2}\right)$.

(B) Given $I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$.
We can rewrite the integral as $I(x)=\int \frac{6 \operatorname{cosec}^2 x}{(1-\cot x)^2} d x$.
Let $t = 1-\cot x$. Then $dt = \operatorname{cosec}^2 x d x$.
Substituting these into the integral,we get $I(x) = \int \frac{6}{t^2} dt = -\frac{6}{t} + C = -\frac{6}{1-\cot x} + C$.
Given $I(0) = 3$. However,$\cot(0)$ is undefined. Assuming the limit as $x \to 0$,$I(x) = \frac{-6}{1-\cot x} + C$.
Actually,evaluating the constant $C$ from the expression $I(x) = \frac{-6}{1-\cot x} + C$,we have $I(x) = \frac{6}{\cot x - 1} + C$.
Given $I(0)=3$ is problematic,but typically such problems imply the constant $C$ is determined by the form. Let's re-evaluate: $I(x) = \frac{6}{\cot x - 1} + C$.
At $x = \frac{\pi}{12}$,$\cot(\frac{\pi}{12}) = \cot(15^\circ) = 2+\sqrt{3}$.
$I(\frac{\pi}{12}) = \frac{6}{2+\sqrt{3}-1} + C = \frac{6}{1+\sqrt{3}} + C = 3(\sqrt{3}-1) + C$.
Assuming $C=3$,$I(\frac{\pi}{12}) = 3\sqrt{3}-3+3 = 3\sqrt{3}$.

(A) Given $f(x) = \frac{x}{(1+x^n)^{1/n}}$.
Then $f(f(x)) = \frac{f(x)}{(1+f(x)^n)^{1/n}} = \frac{x/(1+x^n)^{1/n}}{(1 + x^n/(1+x^n))^{1/n}} = \frac{x}{(1+x^n+x^n)^{1/n}} = \frac{x}{(1+2x^n)^{1/n}}$.
By induction,$g(x) = (f \circ f \circ \ldots \circ f)(x) = \frac{x}{(1+nx^n)^{1/n}}$.
Now,we need to evaluate $I = \int x^{n-2} g(x) \, dx = \int \frac{x^{n-1}}{(1+nx^n)^{1/n}} \, dx$.
Let $u = 1+nx^n$,then $du = n^2 x^{n-1} \, dx$,which implies $x^{n-1} \, dx = \frac{du}{n^2}$.
Substituting these into the integral:
$I = \int \frac{1}{u^{1/n}} \cdot \frac{du}{n^2} = \frac{1}{n^2} \int u^{-1/n} \, du$.
$I = \frac{1}{n^2} \cdot \frac{u^{1 - 1/n}}{1 - 1/n} + K = \frac{1}{n^2} \cdot \frac{u^{(n-1)/n}}{(n-1)/n} + K$.
$I = \frac{1}{n^2} \cdot \frac{n}{n-1} u^{(n-1)/n} + K = \frac{1}{n(n-1)} (1+nx^n)^{1 - 1/n} + K$.

(A) Let $t = \sec x + \tan x$. Then $dt = (\sec x \tan x + \sec^2 x) dx = \sec x(\tan x + \sec x) dx = \sec x \cdot t \cdot dx$.
Thus,$\sec x dx = \frac{dt}{t}$.
Also,$\sec x - \tan x = \frac{1}{t}$.
Adding the two equations: $2 \sec x = t + \frac{1}{t} \implies \sec x = \frac{1}{2} \left( t + \frac{1}{t} \right)$.
Substituting into the integral:
$I = \int \frac{\sec x \cdot (\sec x dx)}{t^{9/2}} = \int \frac{\frac{1}{2}(t + \frac{1}{t}) \cdot \frac{dt}{t}}{t^{9/2}} = \frac{1}{2} \int \frac{t + t^{-1}}{t^{11/2}} dt = \frac{1}{2} \int (t^{-9/2} + t^{-13/2}) dt$.
Integrating:
$I = \frac{1}{2} \left[ \frac{t^{-7/2}}{-7/2} + \frac{t^{-11/2}}{-11/2} \right] + K = -\left[ \frac{1}{7 t^{7/2}} + \frac{1}{11 t^{11/2}} \right] + K$.
Factoring out $-\frac{1}{t^{11/2}}$:
$I = -\frac{1}{t^{11/2}} \left[ \frac{t^2}{7} + \frac{1}{11} \right] + K$.
Substituting $t = \sec x + \tan x$ back,we get option $A$.

(A) Let $x = t^4$,then $dx = 4t^3 dt$.
Substituting these into the integral:
$\int \frac{1}{t(1+t)} \cdot 4t^3 dt = \int \frac{4t^2}{1+t} dt$.
Using polynomial division or adjustment:
$\int \frac{4(t^2-1+1)}{t+1} dt = 4 \int (t-1 + \frac{1}{t+1}) dt$.
Integrating term by term:
$f(x) = 4(\frac{t^2}{2} - t + \log_e|t+1|) + C = 2t^2 - 4t + 4\log_e(t+1) + C$.
Substituting $t = x^{1/4}$:
$f(x) = 2x^{1/2} - 4x^{1/4} + 4\log_e(1+x^{1/4}) + C$.
Given $f(0) = -6$:
$2(0) - 4(0) + 4\log_e(1) + C = -6 \implies C = -6$.
Thus,$f(x) = 2x^{1/2} - 4x^{1/4} + 4\log_e(1+x^{1/4}) - 6$.
Calculating $f(1)$:
$f(1) = 2(1) - 4(1) + 4\log_e(2) - 6 = 2 - 4 + 4\log_e 2 - 6 = 4\log_e 2 - 8 = 4(\log_e 2 - 2)$.

(D) Let $3-x^2 = t^2$. Then $-2x dx = 2t dt$,which implies $x dx = -t dt$.
Also,$x^2 = 3-t^2$.
Substituting these into the integral:
$f(x) = \int (3-t^2) \cdot t \cdot (-t dt) + C$
$f(x) = \int (t^4 - 3t^2) dt + C$
$f(x) = \frac{t^5}{5} - t^3 + C$
Substituting back $t = \sqrt{3-x^2}$:
$f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2} + C$
Given $5f(\sqrt{2}) = -4$,we calculate $f(\sqrt{2})$:
$f(\sqrt{2}) = \frac{(3-2)^{5/2}}{5} - (3-2)^{3/2} + C = \frac{1}{5} - 1 + C = -\frac{4}{5} + C$.
Since $5f(\sqrt{2}) = -4$,we have $5(-\frac{4}{5} + C) = -4$,which gives $-4 + 5C = -4$,so $C = 0$.
Thus,$f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2}$.
Now,calculate $f(1)$:
$f(1) = \frac{(3-1)^{5/2}}{5} - (3-1)^{3/2} = \frac{2^{5/2}}{5} - 2^{3/2} = \frac{4\sqrt{2}}{5} - 2\sqrt{2} = \sqrt{2}(\frac{4}{5} - 2) = \sqrt{2}(\frac{4-10}{5}) = -\frac{6\sqrt{2}}{5}$.

(A) Let $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$.
We can rewrite the numerator as $\sin x = \sin ((x-\alpha) + \alpha)$.
Using the identity $\sin (A+B) = \sin A \cos B + \cos A \sin B$,we get:
$\sin ((x-\alpha) + \alpha) = \sin (x-\alpha) \cos \alpha + \cos (x-\alpha) \sin \alpha$.
Substituting this into the integral:
$I = \int \frac{\sin (x-\alpha) \cos \alpha + \cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} dx$.
$I = \int \cos \alpha dx + \int \sin \alpha \frac{\cos (x-\alpha)}{\sin (x-\alpha)} dx$.
$I = \cos \alpha \int dx + \sin \alpha \int \cot (x-\alpha) dx$.
Integrating,we get $I = x \cos \alpha + \sin \alpha \log |\sin (x-\alpha)| + c$.
Comparing this with the given form $Ax + B \log |\sin (x-\alpha)| + c$,we find $A = \cos \alpha$ and $B = \sin \alpha$.

(B) Let $I = \int \frac{\sin 2x}{(a+b \cos x)^2} dx = \int \frac{2 \sin x \cos x}{(a+b \cos x)^2} dx$.
Substitute $t = a + b \cos x$,then $dt = -b \sin x dx$,so $\sin x dx = -\frac{dt}{b}$.
Also,$\cos x = \frac{t-a}{b}$.
Substituting these into the integral:
$I = \int \frac{2 (\frac{t-a}{b})}{t^2} (-\frac{dt}{b}) = -\frac{2}{b^2} \int \frac{t-a}{t^2} dt$.
$I = -\frac{2}{b^2} \int (\frac{1}{t} - \frac{a}{t^2}) dt$.
$I = -\frac{2}{b^2} [\log |t| + \frac{a}{t}] + C$.
Substituting $t = a + b \cos x$ back:
$I = -\frac{2}{b^2} [\log |a+b \cos x| + \frac{a}{a+b \cos x}] + C$.

(A) Let $I = \int \sin^5 x \, dx = \int \sin^4 x \cdot \sin x \, dx$.
We can write $\sin^4 x = (\sin^2 x)^2 = (1 - \cos^2 x)^2$.
So,$I = \int (1 - \cos^2 x)^2 \sin x \, dx$.
Let $u = \cos x$,then $du = -\sin x \, dx$,which implies $\sin x \, dx = -du$.
Substituting these into the integral:
$I = \int (1 - u^2)^2 (-du) = -\int (1 - 2u^2 + u^4) \, du$.
Integrating term by term:
$I = -(u - \frac{2u^3}{3} + \frac{u^5}{5}) + c = -u + \frac{2}{3} u^3 - \frac{1}{5} u^5 + c$.
Substituting $u = \cos x$ back:
$I = -\cos x + \frac{2}{3} \cos^3 x - \frac{1}{5} \cos^5 x + c$.

(A) Let $I = \int \frac{e^{\tan ^{-1} 2 x}}{1+4 x^2} dx$.
Substitute $u = \tan ^{-1} 2 x$.
Then,$du = \frac{1}{1+(2x)^2} \cdot \frac{d}{dx}(2x) dx = \frac{2}{1+4x^2} dx$.
This implies $\frac{dx}{1+4x^2} = \frac{1}{2} du$.
Substituting these into the integral,we get:
$I = \int e^u \cdot \frac{1}{2} du = \frac{1}{2} \int e^u du$.
$I = \frac{1}{2} e^u + c$.
Substituting $u = \tan ^{-1} 2 x$ back,we get:
$I = \frac{1}{2} e^{\tan ^{-1} 2 x} + c$.

$(A)$ Let $I = \int \frac{x^3}{(x+1)^2} dx$.
Substitute $u = x+1$,then $x = u-1$ and $dx = du$.
$I = \int \frac{(u-1)^3}{u^2} du = \int \frac{u^3 - 3u^2 + 3u - 1}{u^2} du$.
$I = \int (u - 3 + \frac{3}{u} - \frac{1}{u^2}) du$.
Integrating term by term:
$I = \frac{u^2}{2} - 3u + 3\log|u| + \frac{1}{u} + c$.
Substituting $u = x+1$ back:
$I = \frac{(x+1)^2}{2} - 3(x+1) + 3\log|x+1| + \frac{1}{x+1} + c$.
$I = \frac{x^2+2x+1}{2} - 3x - 3 + 3\log|x+1| + \frac{1}{x+1} + c$.
$I = \frac{x^2}{2} + x + \frac{1}{2} - 3x - 3 + 3\log|x+1| + \frac{1}{x+1} + c$.
$I = \frac{x^2}{2} - 2x + 3\log|x+1| + \frac{1}{x+1} + C$ (where $C = c - 2.5$ is a constant).
Thus,the correct option is $A$.

(D) Let $I = \int \frac{\sin x}{\sqrt{5 \sin ^2 x+6 \cos ^2 x}} \,d x$.
Using $\sin ^2 x = 1 - \cos ^2 x$,we get:
$I = \int \frac{\sin x}{\sqrt{5(1 - \cos ^2 x) + 6 \cos ^2 x}} \,d x$
$I = \int \frac{\sin x}{\sqrt{5 - 5 \cos ^2 x + 6 \cos ^2 x}} \,d x$
$I = \int \frac{\sin x}{\sqrt{5 + \cos ^2 x}} \,d x$.
Let $u = \cos x$,then $du = -\sin x \,d x$,which implies $\sin x \,d x = -du$.
Substituting these into the integral:
$I = \int \frac{-du}{\sqrt{5 + u^2}} = -\int \frac{du}{\sqrt{(\sqrt{5})^2 + u^2}}$.
Using the standard formula $\int \frac{dx}{\sqrt{a^2 + x^2}} = \log |x + \sqrt{a^2 + x^2}| + c$:
$I = -\log |u + \sqrt{5 + u^2}| + c$.
Substituting $u = \cos x$ back:
$I = -\log |\cos x + \sqrt{5 + \cos ^2 x}| + c$.

(B) Let $I = \int \frac{\sin 2x \cos 2x}{\sqrt{4-\cos^4 2x}} \, dx$.
Substitute $u = \cos^2 2x$.
Then $du = 2 \cos 2x (-\sin 2x) \cdot 2 \, dx = -4 \sin 2x \cos 2x \, dx$.
So,$\sin 2x \cos 2x \, dx = -\frac{1}{4} du$.
The integral becomes $I = \int \frac{-\frac{1}{4} du}{\sqrt{4-u^2}} = -\frac{1}{4} \int \frac{du}{\sqrt{2^2-u^2}}$.
Using the standard formula $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we get:
$I = -\frac{1}{4} \sin^{-1}\left(\frac{u}{2}\right) + c$.
Substituting back $u = \cos^2 2x$,we obtain:
$I = -\frac{1}{4} \sin^{-1}\left(\frac{\cos^2 2x}{2}\right) + c$.

(A) Let $I = \int \sec^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \, dx$.
We can rewrite the integrand as:
$I = \int \frac{1}{\cos^{\frac{2}{3}} x} \cdot \frac{1}{\sin^{\frac{4}{3}} x} \, dx$.
Divide the numerator and denominator by $\cos^{\frac{4}{3}} x$:
$I = \int \frac{1}{\cos^{\frac{2}{3}} x \cdot \sin^{\frac{4}{3}} x} \cdot \frac{\cos^{\frac{4}{3}} x}{\cos^{\frac{4}{3}} x} \, dx = \int \frac{\cos^{\frac{4}{3}} x}{\cos^2 x \cdot \sin^{\frac{4}{3}} x} \, dx$.
$I = \int \frac{1}{\cos^2 x} \cdot \left( \frac{\sin x}{\cos x} \right)^{-\frac{4}{3}} \, dx = \int \sec^2 x \cdot (\tan x)^{-\frac{4}{3}} \, dx$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
$I = \int u^{-\frac{4}{3}} \, du = \frac{u^{-\frac{4}{3} + 1}}{-\frac{4}{3} + 1} + c = \frac{u^{-\frac{1}{3}}}{-\frac{1}{3}} + c = -3u^{-\frac{1}{3}} + c$.
Substituting $u = \tan x$ back,we get:
$I = -3 \tan^{-\frac{1}{3}} x + c$.

(B) Let $I = \int \frac{d x}{e^x-1}$.
Multiply the numerator and denominator by $e^{-x}$:
$I = \int \frac{e^{-x}}{1-e^{-x}} d x$.
Let $u = 1-e^{-x}$. Then $du = e^{-x} d x$.
Substituting these into the integral:
$I = \int \frac{1}{u} d u = \log |u| + c = \log |1-e^{-x}| + c$.
Since $1-e^{-x} = \frac{e^x-1}{e^x}$,we have:
$I = \log \left| \frac{e^x-1}{e^x} \right| + c = \log |e^x-1| - \log |e^x| + c$.
$I = \log |e^x-1| - x + c$.

(D) To solve the integral $I = \int \frac{dx}{\sqrt{x} + x}$,we can simplify the denominator by factoring out $\sqrt{x}$.
$I = \int \frac{dx}{\sqrt{x}(1 + \sqrt{x})}$
Let $u = 1 + \sqrt{x}$. Then,$du = \frac{1}{2\sqrt{x}} dx$,which implies $\frac{dx}{\sqrt{x}} = 2 du$.
Substituting these into the integral,we get:
$I = \int \frac{2 du}{u} = 2 \int \frac{du}{u} = 2 \log |u| + c$.
Substituting back $u = 1 + \sqrt{x}$,we obtain:
$I = 2 \log (1 + \sqrt{x}) + c$.

(A) To evaluate the integral $I = \int \frac{x}{1+x^4} \, dx$,we can use the method of substitution.
Let $u = x^2$. Then,the derivative is $du = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, du$.
Substituting these into the integral,we get:
$I = \int \frac{1}{1+(x^2)^2} \cdot (x \, dx) = \int \frac{1}{1+u^2} \cdot \frac{1}{2} \, du$
$I = \frac{1}{2} \int \frac{1}{1+u^2} \, du$
Using the standard integral formula $\int \frac{1}{1+u^2} \, du = \tan^{-1}(u) + c$,we have:
$I = \frac{1}{2} \tan^{-1}(u) + c$
Substituting $u = x^2$ back into the expression,we get:
$I = \frac{1}{2} \tan^{-1}(x^2) + c$
Thus,the correct option is $A$.

(B) Let $I = \int \frac{(5 \sin \theta-2) \cos \theta}{5-\cos ^2 \theta-4 \sin \theta} d \theta$.
Substitute $u = \sin \theta$,then $du = \cos \theta d \theta$.
The denominator becomes $5 - (1 - \sin^2 \theta) - 4 \sin \theta = 5 - 1 + u^2 - 4u = u^2 - 4u + 4 = (u-2)^2$.
Thus,$I = \int \frac{5u-2}{(u-2)^2} du$.
Let $u-2 = t$,then $u = t+2$ and $du = dt$.
$I = \int \frac{5(t+2)-2}{t^2} dt = \int \frac{5t+8}{t^2} dt = \int (\frac{5}{t} + 8t^{-2}) dt$.
$I = 5 \log |t| - \frac{8}{t} + c$.
Substituting back $t = u-2 = \sin \theta - 2$,we get $I = 5 \log |\sin \theta - 2| - \frac{8}{\sin \theta - 2} + c$.

(B) To evaluate the integral $I = \int \frac{1}{e^x + 1} \, dx$,we can multiply the numerator and the denominator by $e^{-x}$:
$I = \int \frac{e^{-x}}{e^{-x}(e^x + 1)} \, dx = \int \frac{e^{-x}}{1 + e^{-x}} \, dx$
Let $u = 1 + e^{-x}$. Then $du = -e^{-x} \, dx$,which implies $e^{-x} \, dx = -du$.
Substituting these into the integral:
$I = \int \frac{-du}{u} = -\log|u| + c$
$I = -\log(1 + e^{-x}) + c$
Since $1 + e^{-x} = 1 + \frac{1}{e^x} = \frac{e^x + 1}{e^x}$,we have:
$I = -\log\left(\frac{e^x + 1}{e^x}\right) + c = -[\log(e^x + 1) - \log(e^x)] + c$
$I = -\log(e^x + 1) + x + c$
Thus,$I = x - \log(e^x + 1) + c$.

(A) Let $I = \int \frac{x^4 \cos \left(\tan ^{-1} x^5\right)}{1+x^{10}} \,d x$.
Substitute $u = \tan^{-1}(x^5)$.
Then,$du = \frac{1}{1+(x^5)^2} \cdot \frac{d}{dx}(x^5) \,dx = \frac{5x^4}{1+x^{10}} \,dx$.
This implies $\frac{x^4}{1+x^{10}} \,dx = \frac{du}{5}$.
Substituting these into the integral,we get:
$I = \int \cos(u) \cdot \frac{du}{5} = \frac{1}{5} \int \cos(u) \,du$.
$I = \frac{1}{5} \sin(u) + c$.
Substituting back $u = \tan^{-1}(x^5)$,we get:
$I = \frac{\sin \left(\tan ^{-1} x^5\right)}{5} + c$.

(C) Let $I = \int \frac{dx}{(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}}}$.
Rewrite the integrand as: $I = \int \frac{dx}{(x+a)^{\frac{9}{7}} \cdot (x+a)^{\frac{5}{7}} \cdot \left(\frac{x-b}{x+a}\right)^{\frac{5}{7}}} = \int \frac{dx}{(x+a)^2 \left(\frac{x-b}{x+a}\right)^{\frac{5}{7}}}$.
Let $t = \frac{x-b}{x+a}$. Then $dt = \frac{(x+a)(1) - (x-b)(1)}{(x+a)^2} dx = \frac{a+b}{(x+a)^2} dx$.
Thus,$\frac{dx}{(x+a)^2} = \frac{dt}{a+b}$.
Substituting these into the integral,we get: $I = \int \frac{1}{t^{\frac{5}{7}}} \cdot \frac{dt}{a+b} = \frac{1}{a+b} \int t^{-\frac{5}{7}} dt$.
Integrating,we get: $I = \frac{1}{a+b} \cdot \frac{t^{-\frac{5}{7}+1}}{-\frac{5}{7}+1} + c = \frac{1}{a+b} \cdot \frac{t^{\frac{2}{7}}}{\frac{2}{7}} + c = \frac{7}{2(a+b)} \left(\frac{x-b}{x+a}\right)^{\frac{2}{7}} + c$.

(B) Let $I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \,dx$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sqrt{\tan x} / \cos^2 x}{(\sin x \cos x) / \cos^2 x} \,dx = \int \frac{\sqrt{\tan x} \cdot \sec^2 x}{\tan x} \,dx$.
$I = \int \frac{\sec^2 x}{\sqrt{\tan x}} \,dx$.
Let $u = \tan x$, then $du = \sec^2 x \,dx$.
Substituting these into the integral:
$I = \int \frac{1}{\sqrt{u}} \,du = \int u^{-1/2} \,du$.
Using the power rule $\int u^n \,du = \frac{u^{n+1}}{n+1} + c$:
$I = \frac{u^{1/2}}{1/2} + c = 2 \sqrt{u} + c$.
Substituting back $u = \tan x$:
$I = 2 \sqrt{\tan x} + c$.

(A) To solve the integral $I = \int \frac{dx}{x^{1/2} + x^{1/3}}$,we use the substitution $x = t^6$,so $dx = 6t^5 dt$.
Substituting these into the integral,we get:
$I = \int \frac{6t^5 dt}{t^3 + t^2} = \int \frac{6t^5}{t^2(t+1)} dt = \int \frac{6t^3}{t+1} dt$.
Using polynomial division,$\frac{t^3}{t+1} = t^2 - t + 1 - \frac{1}{t+1}$.
Thus,$I = 6 \int (t^2 - t + 1 - \frac{1}{t+1}) dt = 6 [\frac{t^3}{3} - \frac{t^2}{2} + t - \log|t+1|] + k$.
$I = 2t^3 - 3t^2 + 6t - 6 \log|t+1| + k$.
Substituting $t = x^{1/6}$ back,we get:
$I = 2(x^{1/6})^3 - 3(x^{1/6})^2 + 6(x^{1/6}) - 6 \log(x^{1/6} + 1) + k$.
$I = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6 \log(x^{1/6} + 1) + k$.
Comparing this with the given form $A x^{1/2} + B x^{1/3} + C x^{1/6} + D \log(x^{1/6} + 1) + k$,we find $A = 2, B = -3, C = 6, D = -6$.

(B) Let $I = \int \frac{e^x}{\sqrt{e^{2x}+4e^x+13}} dx$.
Substitute $u = e^x$,then $du = e^x dx$.
The integral becomes $I = \int \frac{du}{\sqrt{u^2+4u+13}}$.
Complete the square for the quadratic expression: $u^2+4u+13 = (u+2)^2 + 9 = (u+2)^2 + 3^2$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^2+a^2}} = \log |x + \sqrt{x^2+a^2}| + c$,we get:
$I = \log |(u+2) + \sqrt{(u+2)^2 + 3^2}| + c$.
Substituting $u = e^x$ back,we get:
$I = \log |e^x + 2 + \sqrt{e^{2x}+4e^x+13}| + c$.
Comparing this with the given expression $\log |e^{ax}+2+\sqrt{e^{2x}+4e^x+13}| + c$,we find that $a = 1$.

(C) Let $I = \int 3^{3^x} \cdot 3^x \, dx$.
Substitute $t = 3^x$.
Then,$dt = 3^x \log 3 \, dx$,which implies $3^x \, dx = \frac{1}{\log 3} \, dt$.
Substituting these into the integral:
$I = \int 3^t \cdot \frac{1}{\log 3} \, dt$.
$I = \frac{1}{\log 3} \int 3^t \, dt$.
Using the formula $\int a^x \, dx = \frac{a^x}{\log a} + c$:
$I = \frac{1}{\log 3} \cdot \frac{3^t}{\log 3} + c$.
$I = \frac{3^t}{(\log 3)^2} + c$.
Substituting $t = 3^x$ back:
$I = \frac{3^{3^x}}{(\log 3)^2} + c$.

(D) Let $I = \int \frac{\log \sqrt{x}}{3 x} \,d x$.
We know that $\log \sqrt{x} = \log (x^{1/2}) = \frac{1}{2} \log x$.
Substituting this into the integral:
$I = \int \frac{\frac{1}{2} \log x}{3 x} \,d x = \frac{1}{6} \int \frac{\log x}{x} \,d x$.
Let $u = \log x$,then $du = \frac{1}{x} \,d x$.
The integral becomes $I = \frac{1}{6} \int u \,du = \frac{1}{6} \cdot \frac{u^2}{2} + c = \frac{u^2}{12} + c$.
Substituting $u = \log x$ back,we get $I = \frac{1}{12} (\log x)^2 + c$.

(B) Let $I = \int \frac{x+1}{\sqrt{2x-1}} \, dx$.
Substitute $2x-1 = t^2$,which implies $x = \frac{t^2+1}{2}$ and $dx = t \, dt$.
Then $x+1 = \frac{t^2+1}{2} + 1 = \frac{t^2+3}{2}$.
Substituting these into the integral:
$I = \int \frac{(\frac{t^2+3}{2}) t \, dt}{t} = \int \frac{t^2+3}{2} \, dt$.
$I = \frac{1}{2} (\frac{t^3}{3} + 3t) + c = \frac{t^3}{6} + \frac{3t}{2} + c$.
Factor out $\frac{t}{6}$:
$I = \frac{t}{6} (t^2 + 9) + c$.
Substitute $t = \sqrt{2x-1}$ back:
$I = \frac{\sqrt{2x-1}}{6} (2x-1+9) + c = \frac{\sqrt{2x-1}}{6} (2x+8) + c$.
$I = \sqrt{2x-1} \cdot \frac{2(x+4)}{6} + c = \sqrt{2x-1} \cdot \frac{x+4}{3} + c$.
Comparing this with $f(x) \sqrt{2x-1} + c$,we get $f(x) = \frac{x+4}{3}$.

(B) Let $I = \int(2x+4)\sqrt{x-1}dx$.
Substitute $t = x-1$,so $x = t+1$ and $dx = dt$.
Then $I = \int(2(t+1)+4)\sqrt{t}dt = \int(2t+6)\sqrt{t}dt$.
$I = \int(2t^{3/2} + 6t^{1/2})dt$.
Integrating term by term:
$I = 2 \cdot \frac{t^{5/2}}{5/2} + 6 \cdot \frac{t^{3/2}}{3/2} + c$.
$I = \frac{4}{5}t^{5/2} + 4t^{3/2} + c$.
Substituting back $t = x-1$:
$I = \frac{4}{5}(x-1)^{5/2} + 4(x-1)^{3/2} + c$.
Comparing this with $a(x-1)^{5/2} + b(x-1)^{3/2} + c$,we get $a = \frac{4}{5}$ and $b = 4$.
Therefore,$2a+b = 2(\frac{4}{5}) + 4 = \frac{8}{5} + 4 = \frac{8+20}{5} = \frac{28}{5}$.

(A) Let $I = \int \frac{\operatorname{cosec} x \, dx}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)}$.
Let $t = 1+\log \left(\tan \frac{x}{2}\right)$.
Differentiating both sides with respect to $x$,we get:
$\frac{dt}{dx} = \frac{1}{\tan \frac{x}{2}} \cdot \sec ^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos ^2 \frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x} = \operatorname{cosec} x$.
Thus,$\operatorname{cosec} x \, dx = dt$.
Substituting these into the integral,we get:
$I = \int \frac{1}{\cos ^2 t} \, dt = \int \sec ^2 t \, dt$.
Integrating,we get $I = \tan t + c$.
Substituting back $t = 1+\log \left(\tan \frac{x}{2}\right)$,we get $I = \tan \left(1+\log \tan \frac{x}{2}\right)+c$.

(D) Let $I = \int \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx$.
We can rewrite the integrand as:
$I = \int \frac{1}{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x} \, dx$.
Divide the numerator and denominator by $\cos^{\frac{4}{3}} x$:
$I = \int \frac{1}{\left(\frac{\sin^{\frac{4}{3}} x}{\cos^{\frac{4}{3}} x}\right) \cdot \cos^{\frac{2}{3}} x \cdot \cos^{\frac{4}{3}} x} \, dx = \int \frac{1}{(\tan x)^{\frac{4}{3}} \cdot \cos^2 x} \, dx$.
Since $\frac{1}{\cos^2 x} = \sec^2 x$,we have:
$I = \int \frac{\sec^2 x}{(\tan x)^{\frac{4}{3}}} \, dx$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$.
Substituting these into the integral:
$I = \int t^{-\frac{4}{3}} \, dt = \frac{t^{-\frac{4}{3} + 1}}{-\frac{4}{3} + 1} + c = \frac{t^{-\frac{1}{3}}}{-\frac{1}{3}} + c = -3t^{-\frac{1}{3}} + c$.
Substituting $t = \tan x$ back:
$I = -3(\tan x)^{-\frac{1}{3}} + c$.

(C) Let $a+b x=t$. Then $x=\frac{t-a}{b}$ and $dx=\frac{dt}{b}$.
Substituting these into the integral:
$I = \int \frac{(\frac{t-a}{b})^2}{t^2} \cdot \frac{dt}{b} = \frac{1}{b^3} \int \frac{t^2-2at+a^2}{t^2} dt$
$I = \frac{1}{b^3} \int (1 - \frac{2a}{t} + \frac{a^2}{t^2}) dt$
$I = \frac{1}{b^3} [t - 2a \log |t| - \frac{a^2}{t}] + c$
Substituting $t=a+bx$ back:
$I = \frac{1}{b^3} [a+bx - 2a \log |a+bx| - \frac{a^2}{a+bx}] + c$